Question

1. The table below shows the joint probability distribution of fatalities (X) and number of seat belts used (Y) for children in accidents. Seat belt status Survivors Fatalities Total No belt 1129 509 1638 Adult belt 432 73 505 Children's carseat belt 733 139 872 Total 2294 3015 721 (a) (1 point) Find E(XY). (b) (1 point) Find Cov(X,Y). (c) (1 point) Find px,y.

Answer 1

X	Y	P(X,Y)	X*Y*P(X,Y)		X*P(X,Y)	(X-E(x))² * P(X,Y)	Y*P(X,Y)	(Y-E(y))² * P(X,Y)
0	0	1129/3015 = 0.3745	0		0.00	0.021	0.00	0.208
0	1	432/3015 = 0.1433	0		0.00	0.008	0.14	0.009
0	2	733/2015 =0.2431	0		0.00	0.014	0.49	0.382
1	0	0.1688	0		0.17	0.098	0.00	0.094
1	1	0.0242	0.0242		0.024212	0.014	0.024212	0.001562854
1	2	0.0461	0.092206		0.046103	0.027	0.092206	0.072504708
	total	1	0.116418		0.24	0.182	0.75	0.768

a)

E(XY)=ΣXYP(XY) =        0.1164

b) E[X] = Σx*P(X) =        0.2391                          
E[Y] = ΣY*P(Y) =        0.7459                          
Cov(X,Y) = E(XY)-E(X)E(Y)=           0.11641791   -   0.24   *   0.746   =   -0.0620

c)

Var(X) =    Σ(X-E(x))² * P(X,Y)=   0.1820   0.42655695                          
Var(Y) =    Σ(Y-E(y))² * P(X,Y)=   0.7680   0.876331061                          
correlation=cov(x,y)/√(Var(x)*Var(y)) =                -0.06   /√(   0.182   *   0.768   )=   -0.1658 (answer)